# Question 3d879

Sep 28, 2017

$x$=$\sqrt{1 - {k}^{2}}$ when 0≦k≦1.

#### Explanation:

In this equation, $x$=$\sin \left\{\arcsin \left(x\right)\right\}$
=$\sin${$\frac{\pi}{2}$-$\arcsin \left(k\right)$}
=$\cos \left\{\arcsin \left(k\right)\right\}$.

By difinition it follows $- \frac{\pi}{2}$$\arcsin \left(k\right)$$\frac{\pi}{2}$ and
$\cos \left\{\arcsin \left(k\right)\right\}$ is non-negative.
So, $x$=$\cos \left\{\arcsin \left(k\right)\right\}$=sqrt(1-sin^2{arcsin(k)}
=$\sqrt{1 - {k}^{2}}$.
You may think domain of $k$ is -1≦k≦1# but this is wrong.

If you solve this equation for $k$ instead of $x$, $k = \sqrt{1 - {x}^{2}}$
and $k$ must be non-negative too.
The true domain is $0 \le k \le 1$.